3t^2-18t+20=0

Solution for 3t^2-18t+20=0 equation:

3t^2-18t+20=0

a = 3; b = -18; c = +20;
Δ = b2-4ac
Δ = -182-4·3·20
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{21}}{2*3}=\frac{18-2\sqrt{21}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{21}}{2*3}=\frac{18+2\sqrt{21}}{6}
$